问题
Let us suppose that a std::tuple<some_types...> is given. I would like to create a new std::tuple whose types are the ones indexed in [0, sizeof...(some_types) - 2]. For instance, let's suppose that the starting tuple is std::tuple<int, double, bool>. I would like to obtain a sub-tuple defined as std::tuple<int, double>.
I'm quite new to variadic templates. As a first step I tried to write a struct in charge of storing the different types of the original std::tuple with the aim of creating a new tuple of the same kind (as in std::tuple<decltype(old_tuple)> new_tuple).
template<typename... types>
struct type_list;
template<typename T, typename... types>
struct type_list<T, types...> : public type_list<types...> {
typedef T type;
};
template<typename T>
struct type_list<T> {
typedef T type;
};
What I would like to do is something like:
std::tuple<type_list<bool, double, int>::type...> new_tuple // this won't work
And the next step would be of discarding the last element in the parameter pack. How can I access the several type's stored in type_list? and how to discard some of them?
Thanks.
回答1:
This kind of manipulation is fairly easy with an index sequence technique: generate an index sequence with two fewer indices than your tuple, and use that sequence to select fields from the original. Using std::make_index_sequence and return type deduction from C++14:
template <typename... T, std::size_t... I>
auto subtuple_(const std::tuple<T...>& t, std::index_sequence<I...>) {
return std::make_tuple(std::get<I>(t)...);
}
template <int Trim, typename... T>
auto subtuple(const std::tuple<T...>& t) {
return subtuple_(t, std::make_index_sequence<sizeof...(T) - Trim>());
}
In C++11:
#include <cstddef> // for std::size_t
template<typename T, T... I>
struct integer_sequence {
using value_type = T;
static constexpr std::size_t size() noexcept {
return sizeof...(I);
}
};
namespace integer_sequence_detail {
template <typename, typename> struct concat;
template <typename T, T... A, T... B>
struct concat<integer_sequence<T, A...>, integer_sequence<T, B...>> {
typedef integer_sequence<T, A..., B...> type;
};
template <typename T, int First, int Count>
struct build_helper {
using type = typename concat<
typename build_helper<T, First, Count/2>::type,
typename build_helper<T, First + Count/2, Count - Count/2>::type
>::type;
};
template <typename T, int First>
struct build_helper<T, First, 1> {
using type = integer_sequence<T, T(First)>;
};
template <typename T, int First>
struct build_helper<T, First, 0> {
using type = integer_sequence<T>;
};
template <typename T, T N>
using builder = typename build_helper<T, 0, N>::type;
} // namespace integer_sequence_detail
template <typename T, T N>
using make_integer_sequence = integer_sequence_detail::builder<T, N>;
template <std::size_t... I>
using index_sequence = integer_sequence<std::size_t, I...>;
template<size_t N>
using make_index_sequence = make_integer_sequence<size_t, N>;
#include <tuple>
template <typename... T, std::size_t... I>
auto subtuple_(const std::tuple<T...>& t, index_sequence<I...>)
-> decltype(std::make_tuple(std::get<I>(t)...))
{
return std::make_tuple(std::get<I>(t)...);
}
template <int Trim, typename... T>
auto subtuple(const std::tuple<T...>& t)
-> decltype(subtuple_(t, make_index_sequence<sizeof...(T) - Trim>()))
{
return subtuple_(t, make_index_sequence<sizeof...(T) - Trim>());
}
Live at Coliru.
回答2:
Here is a way to solve your problem directly.
template<unsigned...s> struct seq { typedef seq<s...> type; };
template<unsigned max, unsigned... s> struct make_seq:make_seq<max-1, max-1, s...> {};
template<unsigned...s> struct make_seq<0, s...>:seq<s...> {};
template<unsigned... s, typename Tuple>
auto extract_tuple( seq<s...>, Tuple& tup ) {
return std::make_tuple( std::get<s>(tup)... );
}
You can use this as follows:
std::tuple< int, double, bool > my_tup;
auto short_tup = extract_tuple( make_seq<2>(), my_tup );
auto skip_2nd = extract_tuple( seq<0,2>(), my_tup );
and use decltype if you need the resulting type.
A completely other approach would be to write append_type, which takes a type and a tuple<...>, and adds that type to the end. Then add to type_list:
template<template<typename...>class target>
struct gather {
typedef typename type_list<types...>::template gather<target>::type parent_result;
typedef typename append< parent_result, T >::type type;
};
which gives you a way to accumulate the types of your type_list into an arbitrary parameter pack holding template. But that isn't required for your problem.
回答3:
One way to do it is to recursively pass two tuples to a helper struct that takes the first element of the "source" tuple and adds it to the end of the another one:
#include <iostream>
#include <tuple>
#include <type_traits>
namespace detail {
template<typename...>
struct truncate;
// this specialization does the majority of the work
template<typename... Head, typename T, typename... Tail>
struct truncate< std::tuple<Head...>, std::tuple<T, Tail...> > {
typedef typename
truncate< std::tuple<Head..., T>, std::tuple<Tail...> >::type type;
};
// this one stops the recursion when there's only
// one element left in the source tuple
template<typename... Head, typename T>
struct truncate< std::tuple<Head...>, std::tuple<T> > {
typedef std::tuple<Head...> type;
};
}
template<typename...>
struct tuple_truncate;
template<typename... Args>
struct tuple_truncate<std::tuple<Args...>> {
// initiate the recursion - we start with an empty tuple,
// with the source tuple on the right
typedef typename detail::truncate< std::tuple<>, std::tuple<Args...> >::type type;
};
int main()
{
typedef typename tuple_truncate< std::tuple<bool, double, int> >::type X;
// test
std::cout << std::is_same<X, std::tuple<bool, double>>::value; // 1, yay
}
Live example.
回答4:
Subrange from tuple with boundary checking, without declaring "helper classes":
template <size_t starting, size_t elems, class tuple, class seq = decltype(std::make_index_sequence<elems>())>
struct sub_range;
template <size_t starting, size_t elems, class ... args, size_t ... indx>
struct sub_range<starting, elems, std::tuple<args...>, std::index_sequence<indx...>>
{
static_assert(elems <= sizeof...(args) - starting, "sub range is out of bounds!");
using tuple = std::tuple<std::tuple_element_t<indx + starting, std::tuple<args...>> ...>;
};
Usage:
struct a0;
...
struct a8;
using range_outer = std::tuple<a0, a1, a2, a3, a4, a5, a6, a7, a8>;
sub_range<2, 3, range_outer>::tuple; //std::tuple<a2, a3, a4>
来源：https://stackoverflow.com/questions/17854219/creating-a-sub-tuple-starting-from-a-stdtuplesome-types